## anonymous one year ago A 2.0 µC point charge travels a distance of 0.20 cm between two parallel charged plates from the negative end toward the positive end. The electric field has a strength of 500.0 N.C. What is the change in voltage? (µC=1.0*10^-6 C)

1. anonymous

A. -100 V B. +1.0 V C. 0 V D. +0.20 mV

2. Michele_Laino

the subsequent relationship holds: $\Delta V = Ed$ |dw:1433273196060:dw|

3. anonymous

ok! so what happens next? :/

4. anonymous

500.0 x 0.2

5. anonymous

=100

6. anonymous

7. anonymous

ohh okay!! thank you!!:D

8. Michele_Laino

no, it is: $\Large \Delta V = Ed = 500 \times 2 \times {10^{ - 3}} = ...$ since the distance is 0.2 cm = 2*10^(-3) meters

9. anonymous

so +1.0V is the solution?

10. Michele_Laino

yes! since our charge is approaching to the positive plane, so its potential is increasing, namlely final potential is larger than initial potential so the difference between them is positive

11. anonymous

yay! thank you both!!

12. Michele_Laino

:)